Feedback To Source

[nievesoliveras]

Hi,There is a very clever and gifted user on utube called "theDaftman".He has made a battery swapper circuit which monitors the voltage and when it reaches a level set by a pot ,it swaps batteries.He is using it to swap 12v batteries but you may be able to adjust it for 1.5v batteries.Here is a link to his video.You can download his circuit, details in vid.
http://www.youtube.com/watch?v=BbQQYEsxDpg
Also you may consider using a 4 pole double throw relay.If you can find one which latches(these have 2 coils and have the advantage that they don't need continous power.)If you can't find a 4pdt latching relay,you may be able to modify an ordinary single coil 4pdt relay by adding another coil or use 2 latching 2pdt relays in parallel.Hope this helps.
                                                      regards jonnydavro

Thank you @jonnydavro

By the way I am a member of his site. But last year he said that he could not share that information because it was not his to share.
Now I remember where I read about that swapper.

I will check the video and try to log in on the site if my account is still active.
I even have an information of the tesla switch on my hardisk. I dont know how I forgot.

Thank you!

Jesus

[Low-Q]

If the person on the video is you. You seem a very inteligent person.
There is a misunderstanding in all this.  If you have been following this thread from the beginning. You will know by now that the purpose I had is to make a pulse motor to run with a 1.5v and recharge the battery while doing so.

Ok. When I screamed eureka I was screaming not because of the newman motor. If you check the circuit again, you will see that the newman motor does nothing to feedback anything to the source.

The newman motor (I will say pulse motor, because any pulse motor works with this circuit) is just the one I had at hand to make experiments with, that do not take too much space.

There is a problem where I live and is that the wind seldom blows.  I bought a wind turbine kit. The wind does not blow. Then I need to device something to be the wind. In this case a pulse motor. The motor will not do any electric generation whatsoever. Who is going to do all the job is the 1.5v runner circuit.

If you take the circuit I posted and eliminate the pulse motor circuit completely and let only its 2200uf capacitor in place. This circuit will (with an old 1.5v battery) charge a pack of 4 1.5v batteries that somehow the charge given to them does not work on a camera to take a photo, but if you put back the pulse motor and run it with that same pack of batteries, it will work.

Dont be my enemy, Be my friend.
We can share information about motors.
I love electromechanics.  I am retired now, but I use to work on an electromechanic shop for ten years.

Again is not the pulse motor what charge the batteries, is the toroid or Joule thief circuit that does the job.

Lets be friends.

Jesus

Hi,

I'm not the person in the video, so the first conclusion is that I'm not that intelligent - 

However, my intelligence tells me that if you're gonna charge a battery from a 1,5VAA cell, you must take all that capacity from that AA cell. Further, this AA cell is running a circuit that enables a motor to run. This motor is then in turn claimed to charge the battery inside the circuit. Well, I don't tell it's impossible. I see that motor as the same as a step up transformer that enables several volt output. That said, this output also have a current flow, measured in ampéres.

So the questions is:
How much Ah (ampere hour) is the battery capable of being charged, and how many Ah is used from the 1,5V AA cell?
Further the question is also about how much voltage both AA cell and battery represent at the end of charging?

Next stop is now the question on how to use the charged battery to use its excess capacity to replace the 1,5V AA cell. If the total capacity in Ah is more on a fully charged battery than you use in Ah from the 1,5V AA cell, you can simply replace the 1,5V AA cell with a small capacitor, where the plus end is connected to plus on the charged battery via a resistor.

Now, with the right size of the resistor, the charged battery will apply 1,5V in the place of the previous 1,5V AA cell (where approx 4,5V is resting over the resistor), with the same current flow for running that motor. So the motor can charge the battery of four AA cells again - as i have understood the motor is essensial to make a proper charge of those four AA cells (the battery) - to run your camera.

Hey, Jesus. I'm not your enemy, and I will gladly share my knowledge about electro motors. And with ten years experience, you sure know how to make a proper motor or a generator for that matter. So keep on going, but be aware that equation: P = U x I (P = watt, U = voltage, I = ampére). P is all that matters.

Just a hint from the sideline, and some off topic: Phase shift between current and voltage, can also trick ones brain to believe there is overunity. Say you have a capacitor and an inductor in parallell. At the resonance frequency, the applied AC voltage does almost not apply current at all. However, the current flown through the capacitor and the inductor internally might be several tens of ampéres, but almost 180 degree out of phase of each other. For sure you're able to measure 10 amps at the same voltage with instruments - making you believe you have much more power out than you put in.

Br.

Vidar

[nievesoliveras]

Hi,

I'm not the person in the video, so the first conclusion is that I'm not that intelligent - 

However, my intelligence tells me that if you're gonna charge a battery from a 1,5VAA cell, you must take all that capacity from that AA cell. Further, this AA cell is running a circuit that enables a motor to run. This motor is then in turn claimed to charge the battery inside the circuit. Well, I don't tell it's impossible. I see that motor as the same as a step up transformer that enables several volt output. That said, this output also have a current flow, measured in ampéres.

So the questions is:
How much Ah (ampere hour) is the battery capable of being charged, and how many Ah is used from the 1,5V AA cell?
Further the question is also about how much voltage both AA cell and battery represent at the end of charging?

Next stop is now the question on how to use the charged battery to use its excess capacity to replace the 1,5V AA cell. If the total capacity in Ah is more on a fully charged battery than you use in Ah from the 1,5V AA cell, you can simply replace the 1,5V AA cell with a small capacitor, where the plus end is connected to plus on the charged battery via a resistor.

Now, with the right size of the resistor, the charged battery will apply 1,5V in the place of the previous 1,5V AA cell (where approx 4,5V is resting over the resistor), with the same current flow for running that motor. So the motor can charge the battery of four AA cells again - as i have understood the motor is essensial to make a proper charge of those four AA cells (the battery) - to run your camera.

Hey, Jesus. I'm not your enemy, and I will gladly share my knowledge about electro motors. And with ten years experience, you sure know how to make a proper motor or a generator for that matter. So keep on going, but be aware that equation: P = U x I (P = watt, U = voltage, I = ampére). P is all that matters.

Just a hint from the sideline, and some off topic: Phase shift between current and voltage, can also trick ones brain to believe there is overunity. Say you have a capacitor and an inductor in parallell. At the resonance frequency, the applied AC voltage does almost not apply current at all. However, the current flown through the capacitor and the inductor internally might be several tens of ampéres, but almost 180 degree out of phase of each other. For sure you're able to measure 10 amps at the same voltage with instruments - making you believe you have much more power out than you put in.

Br.

Vidar

Thank you @lowq

What I did now is I disconnected the pulse motor by pulling the trigger negative wire and am testing the circuit to see what happens.

I have somewhere on my hardisk a schematic of a way to increase the amps when the voltage is ac.
I will fetch it and post it for your consideration. By the way I dont know where I found it. I have been years on my energy quest.

Jesus

[nievesoliveras]

@lowq

I was lucky today. It was on the first directory searched.
According with the schematic heading it seems that I found it on a search done at EnergeticForum.
Instead of 120v. The output of a toroid coil could be used in order to get feedback to the source that has more amps.

I could be wrong though.

Jesus

[Pirate88179]

@ Jonnydavro:

That is exactly what I envisioned and what I posted about a few pages back.  Fantastic!  I knew there should be a way to do that, I just didn't know how.  This is very good information.

Bill